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7/8/2013
1
TOPIC 4: PUMPS AND TURBINES
OBJECTIVES
1. Understand the role of pumps and turbines as energy-conversion devices
and use, appropriately, the terms head, power and efficiency.
2. Be aware of the main types of pumps and turbines and the distinction
between impulse and reaction turbines and between radial, axial and
mixed-flow devices.
3. Match pump characteristics and system characteristics to determine the
duty point.
4. Calculate characteristics for pumps in series and parallel and use the
hydraulic scaling laws to calculate pump characteristics at different
speeds.
5. Select the type of pump or turbine on the basis of specific speed.
6. Understand the mechanics of a centrifugal pump and an impulse turbine.
7. Recognise the problem of cavitation and how it can be avoided.
ENERGY CONVERSION
Pumps: electrical/mechanical energy fluid energy
Turbines: fluid energy electrical/mechanical energy
PUMP
pipeline
TURBINE
total head
total head
pipeline
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ENERGY
Energy per unit weight = head, H
Power = rate of conversion of energy
Efficiency
Turbines:
Pumps:
g
Vz
g
pH
2ρ
2
gQHρpower(fluid)
in
out
power
powerη
gQHρη outpower
inpower
gQHρη
EXAMPLE, PAGE 2
A pump lifts water from a large tank at a rate of 30 L s–1.
If the input power is 10 kW and the pump is operating at
an efficiency of 40%, find:
(a) the head developed across the pump;
(b) the maximum height to which it can raise water if the
delivery pipe is vertical, with diameter 100 mm and
friction factor λ = 0.015.
TYPES OF PUMPS AND TURBINES:
Impulse vs Reaction Turbines
Impulse turbine (Pelton wheel; water wheel)
– change in velocity change in head
Reaction turbine (Francis turbine, Kaplan turbine, windmill)
– change in pressure change in head
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TYPES OF PUMPS AND TURBINES:
Positive-Displacement vs Dynamic Pumps
Positive-displacement (heart; bicycle pump; peristaltic pump)
– change in volume
Dynamic pump
– no change of volume; continuous transfer of energy
– commonest are rotodynamic pumps
TYPES OF PUMPS AND TURBINES:
Radial, Axial and Mixed-Flow Devices
high head,
low flow
low head,
high flow
Radial Axial Mixed
PUMPS
Centrifugal pump
Axial-flow pump
Inward-flow reaction turbine centrifugal pump
(e.g. Francis turbine)
Propeller turbine axial-flow pump
(e.g. Kaplan turbine; windmill)
Volute
'Eye' (intake)
Impeller vane
flow
rotation
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TURBINES
Pelton wheel
– very-high head; hydropower
Francis turbine
– intermediate head; hydropower (pumped storage)
Kaplan turbine
– axial flow
Wells turbine
– wave energy
Bulb turbine
– tidal power
Archimedean screw
– small-scale hydro-power
PELTON WHEEL
PELTON WHEEL
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FRANCIS TURBINE
KAPLAN TURBINE
ARCHIMEDEAN SCREW
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PUMP CHARACTERISTICS
Head: H vs Q
Input power: I vs Q
Efficiency: η vs Q
Q
I
H
Q
IH
centrifugal pump axial-flow pump
PUMP CHARACTERISTICS:
Lab Results
0.0
2.0
4.0
6.0
8.0
10.0
12.0
0 50 100 150
H (m
)
Q (L min-1)
1.1 Head vs Discharge (2800 rpm)
0
50
100
150
200
250
300
350
0 50 100 150
I (W
)
Q (L min-1)
1.2 Input Power vs Discharge (2800 rpm)
0
5
10
15
20
25
30
35
40
45
50
0 50 100 150
(%
)
Q (L min-1)
1.3 Efficiency vs Discharge (2800 rpm)
SYSTEM CHARACTERISTICS
The pump is required to:
lift water through a certain height (static lift, Hs)
overcome frictional (and other flow-related) losses
2aQHH s
Sump
Delivery reservoir
Suction main
Delivery mainStatic liftHs
Suction headPump
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DUTY POINT
The discharge Q is such that … the head provided by the
pump exactly matches the system head (static lift + losses)
The corresponding Q and H are called the duty point.
Ideally, this should be close to the maximum-efficiency point.
Q
HSystemcharacteristic
Pumpcharacteristic
Dutypoint
Hs
EXAMPLE, PAGE 7
A water pump was tested at a rotation rate of 1500 rpm. The following data was
obtained. (Q is quantity of flow, H is head of water, η is efficiency).
It is proposed to used this pump to draw water from an open sump to an
elevation 5.5 m above. The delivery pipe is 20.0 m long and 100 mm diameter,
and has a friction factor of 0.005.
If operating at 1500 rpm, find:
(a) the maximum discharge that the pump can provide;
(b) the pump efficiency at this discharge;
(c) the input power required.
Q (L s–1) 0 10 20 30 40 50
H (m) 10.0 10.5 10.0 8.5 6.0 2.5
η 0.0 0.40 0.64 0.72 0.64 0.40
PUMPS IN PARALLEL AND SERIES
Parallel
Same head: H
Add discharges: Q1 + Q2
Series
Same discharge: Q
Add heads: H1 + H2
Q
H
Single pump
Double the flow
Pumps in parallel
Q
H
Single pump
Pumps in series
Doublethe
head
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PUMPS IN PARALLEL AND SERIES:
Lab Results
0.0
5.0
10.0
15.0
20.0
0 50 100 150
H (m
)
Q (L min-1)
3.1 Head vs Discharge (Series)
Measured
Scaled
EXAMPLE, PAGE 9A rotodynamic pump, having the characteristics tabulated below, delivers water
from a river at elevation 102 m to a reservoir with a water level of 135 m, through a
350 mm diameter cast-iron pipe. The frictional head loss in the pipeline is given by
hf = 550 Q2, where hf is the head loss in m and Q is the discharge in m3 s–1. Minor
head losses from valves and fittings amount to 50 Q2 in the same units.
(a) Calculate the discharge and head in the pipeline (at the duty point).
If the discharge is to be increased by the installation of a second identical pump:
(b) determine the unregulated discharge and head produced by connecting the
pump:
(i) in parallel;
(ii) in series;
(c) determine the power demand at the duty point in the case of parallel operation.
Q (m3 s–1) 0 0.05 0.10 0.15 0.20
H (m) 60 58 52 41 25
η (%) --- 44 65 64 48
DIMENSIONAL ANALYSIS
Variables and dimensions:
discharge Q [L3T–1]
pressure change ρgH [ML–1T–2]
power P [ML2T–3]
rotor diameter D [L]
rotation rate N [T–1]
fluid density ρ [ML–3]
fluid viscosity μ [ML–1T–1]
# variables = 7
# dimensions = 3 (M,L,T)
# dimensionless Π groups = 4
31ΠND
Q
222ΠDN
gH 533
ρΠ
DN
P Re
μ
ρΠ
2
4 ND
discharge head power viscosity
Scales: D, N, ρ
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DIMENSIONAL ANALYSIS
31ΠND
Q
222ΠDN
gH 533
ρΠ
DN
P Re
μ
ρΠ
2
4 ND
discharge head power viscosity
For fully-turbulent flow there is no significant dependence on Π4 (Re)
Any one of Π1, Π2, Π3 can be replaced by:
or the reciprocal of this for turbines
ηρ
Π
ΠΠ
3
21 P
gQHefficiency (for pumps)
EXAMPLE, PAGE 10
A ¼-scale model centrifugal pump is tested under a
head of 7.5 m at a speed of 500 rpm. It was found that
7.5 kW was needed to drive the model. Assuming similar
mechanical efficiencies, calculate:
(a) the speed and power required by the prototype when
pumping against a head of 44 m;
(b) the ratio of the discharges in the model to that in the
prototype.
HYDRAULIC SCALING LAWS:
(“Affinity Laws”)
2
3
1
3
ND
Q
ND
Q
2
22
1
22
DN
gH
DN
gH
2
53
1
53 ρρ
DN
P
DN
P
Speed
For the same pump/turbine and working fluid (same D, ρ):
1
2
1
2
N
N
Q
Q
2
1
2
1
2
N
N
H
H3
1
2
1
2
N
N
P
P21 ηη
discharge speed head speed2 power speed3
Size
For different-sized, but geometrically similar devices at same speed (same N, ρ):
21 ηη 3
1
2
1
2
D
D
Q
Q2
1
2
1
2
D
D
H
H5
1
2
1
2
D
D
P
P
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FINDING THE DUTY POINT AT A NEW PUMP
SPEED
Hydraulic scaling:1
2
1
2
N
N
Q
Q
2
1
2
1
2
N
N
H
H
Q
System characteristic
New duty point
Hs
N1
N2
H
FINDING THE PUMP SPEED FOR A GIVEN
DUTY POINT
The unknown speed N2 can be found from the ratio of discharges or ratio of heads:
2
22
Q
Q
H
H
Plot a hydraulic-scaling curve back from
the required duty point (Q2,H2) on the
system curve at unknown speed N2:
This will cut the given curve at point (Q1,H1).
1
2
1
2
Q
Q
N
N
1
2
2
1
2
H
H
N
N
or
Q
System characteristic
New duty point
Hs
N1
N2
Scaling curvethroughduty point
H
),( 22 HQ
),( 11 HQ
EXAMPLE, PAGE 12
Water from a well is pumped by a centrifugal pump which delivers water to a
reservoir in which the water level is 15.0 m above that in the sump. When the
pump speed is 1200 rpm its pipework has the following characteristics:
Pipework characteristics:
Discharge (L s–1): 20 30 40 50 60
Head loss in pipework (m): 1.38 3.11 5.52 8.63 12.40
Pump characteristics:
Discharge (L s–1): 0 10 20 30 40
Head (m): 22.0 21.5 20.4 19.0 17.4
A variable-speed motor drives the pump.
(a) Plot the graphs of the system and pump characteristics and determine the
discharge at a speed of 1200 rpm.
(b) Find the pump speed in rpm if the discharge is increased to 40 L s–1.
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SPECIFIC SPEED (PUMPS)
31ΠND
Q
222ΠDN
gH
Find a combination independent of D ...3
42
3
2
2
1
)()Π(
)Π(
gH
NQ
... and proportional to N:4/3
2/14/1
3
2
2
1
)(Π
Π
gH
NQ
Since g is constant, it follows that, as speed changes,
4/3
2/1
H
NQ
is the same at any given (e.g. maximum) efficiency.
Most important parameters are discharge and head:
SPECIFIC SPEED (PUMPS)
4/3
2/1
H
NQNs
Specific speed is that rotation rate which will generate 1 unit of head for 1 unit
of discharge:
Single value, calculated at maximum-efficiency point.
Usual units (UK): N in rpm, Q in m3 s–1, H in m.
Less-common alternative: dimensionless specific speed,
Typical ranges:
4/3
2/1
)(gH
NQKn
Type Ns
Centrifugal 10 – 70 large head
Mixed flow 70 – 170
Axial > 110 small head
EXAMPLE, PAGE 14
A pump is needed to operate at 3000 rpm (i.e.
50 Hz) with a head of 6 m and a discharge of
0.2 m3 s–1. By calculating the specific speed,
determine what sort of pump is required.
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SPECIFIC SPEED (TURBINES)
222ΠDN
gH
Find a combination independent of D ... 5
42
5
2
2
3
)(
)ρ/(
)Π(
)Π(
gH
NP
... and proportional to N:
Since ρ and g are constant, it follows that, as speed changes,
is the same at any given (e.g. maximum) efficiency.
533ρ
ΠDN
P
4/5
2/1
H
NPNS
Most important parameters are head and (output) power:
4/5
2/14/1
5
2
2
3
)(
)ρ/(
Π
Π
gH
NP
SPECIFIC SPEED (TURBINES)
Specific speed is that rotation rate which will generate 1 unit of power for 1 unit
of head:
Single value, calculated at maximum-efficiency point.
Usual units (UK): N in rpm, P in kW, H in m.
Less-common alternative: dimensionless specific speed,
Typical ranges:
4/5
2/1
H
NPN S
4/5
2/1
)(
)ρ/(
gH
NPKn
Type Ns
Pelton wheel (impulse) 12 – 60 very large head
Francis turbine (radial) 60 – 500 large head
Kaplan turbine (axial-flow) 280 – 800 small head
MECHANICS OF CENTRIFUGAL PUMPS
Volute
'Eye' (intake)
Impeller vane
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MECHANICS OF CENTRIFUGAL PUMPS
u impeller velocity (u = rω)
w velocity relative to the impeller
v = u + w absolute velocity
Radial velocity vr determined by flow rate Q
Whirl velocity vt is a combination of impeller (rω) and tangential velocity relative to vanes
2w2β
1w
ω22 ru
ω11 ru
Vane
v,Resultant
w β
ωru
v
tv
rv
MECHANICS OF CENTRIFUGAL PUMPS
Torque = rate of change of angular momentum
)(ρ 1122 rvrvQT tt
Power = torque angular velocity
ω)(ρρ 1122 rvrvQgQH tt
Impeller velocity u = rω
Euler’s turbomachinery equation:
)(1
1122 uvuvg
H tt
Vane
v,Resultant
w β
ωru
v
tv
rv
EFFECT OF BLADE ANGLE
Usually design for small inlet whirl velocity: vt1 0.
22
1uv
gH tThen
βsinwA
Qvr
βcoswuvt
)βcot( 22
A
Qu
g
uH
bQaH
Vane
v,Resultant
w β
ωru
v
tv
rv
Q
H
90β
90β
90β
(backward-facing blades)
(forward-facing blades)
7/8/2013
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EXAMPLE, PAGE 17
A centrifugal pump is required to provide a head of
40 m. The impeller has outlet diameter 0.5 m and
inlet diameter 0.25 m and rotates at 1500 rpm. The
flow approaches the impeller radially at 10 m s–1
and the radial velocity falls off as the reciprocal of
the radius. Calculate the required vane angle at the
outlet of the impeller.
MECHANICS OF A PELTON WHEEL
Force (on fluid) = mass flux change in velocity
Change in velocity easiest to establish in the frame of the bucket
)θcos1)((
)(θcos)(
kuv
uvuvkx
velocity- in Change
)θcos1)((ρ kuvQF
Power = force (on bucket) velocity (of bucket) = Fu
)θcos1()(ρ kuuvQP (per jet)
JetBucket
Spear valve
bucket
jet uv
k(v-u)
v-u
MECHANICS OF A PELTON WHEEL
)θcos1()(ρ kuuvQP
2
212
21 )()()( uvvuuv
vu21
Power (per jet)
Theoretical maximum power when
runner speed equals half the jet speed:
If θ = 180 and k = 1 this corresponds
to the absolute velocity on exit being 0
and all kinetic energy converted.
In practice:
θ 165° (to avoid interference with the following bucket)
the jet velocity v is controlled; (turbine usually synchronised to the electricity grid)
the optimal speed ratio u / v is nearer 0.46.
Other considerations:
Head at device, H = (head at reservoir) − (head losses along pipeline)
Maximum jet velocity,
Orifice coefficient, cv 0.97 – 0.99
gHcv v 2
bucket
jet uv
k(v-u)
v-u
7/8/2013
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EXAMPLE, PAGE 19
In a Pelton wheel, 6 jets of water, each with a
diameter of 75 mm and carrying a discharge of
0.15 m3 s–1 impinge on buckets arranged around a
1.5 m diameter Pelton wheel rotating at 180 rpm.
The water is turned through 165° by each bucket
and leaves with 90% of the original relative velocity.
Neglecting mechanical and electrical losses within
the turbine, calculate the power output.
CAVITATION
Bubbles of vapour form if pressure is too low; a problem on:
– inlet / suction side of pump
– outlet / draft tube of turbine
Subsequent bubble collapse causes:
– surface pitting
– vibration
– noise
– loss of performance
CAVITATION
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NET POSITIVE SUCTION HEAD
g
pp cavinlet
ρNPSH
losshead sumpinletpump HH
Bernoulli:
fatm
inletinlet h
g
p
g
Vz
g
p
ρ2ρ
2
finletatminlet h
g
Vz
g
p
g
p
2ρρ
2
To prevent cavitation, keep pinlet as large as possible by:
keeping zinlet small (better still, negative – i.e. below level in sump)
keeping V small (use large-diameter pipes)
keeping hf small (use short, large-diameter pipes)
Keeping the pump below the level in the sump is also useful for pump-priming.
Sump
pump inlet
zinletpatm
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